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\title{\heiti\zihao{2} 习题6.2}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{应用换元积分法求下列不定积分:}
\subsection{$\int e^{2 x+1} \mathrm{~d} x$}
\textbf{解}\quad
$$\int e^{2 x+1} \mathrm{~d} x=\frac{1}{2} \int e^{2 x+1} \mathrm{~d} (2x+1)=\frac{1}{2} e^{2 x+1}+C$$


\subsection{$\int \frac{1}{3 x+2} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{1}{3 x+2} \mathrm{~d} x=\frac{1}{3} \int \frac{1}{3 x+2} \mathrm{~d} (3 x+2)=\frac{1}{3} \ln |3 x+2|+C$$


\subsection{$\int \frac{1}{1-\sin x} \mathrm{~d} x$}
\textbf{解}\quad
令$u=\tan \frac{x}{2}$,则$\mathrm{~d} x=\frac{2}{1+u^2}\mathrm{~d} u$

$$\int \frac{1}{1-\sin x} \mathrm{~d} x=\int \frac{2}{1+u^{2}-2 u} \mathrm{~d} u=\int \frac{2}{(u-1)^{2}} \mathrm{~d} u=\frac{-2}{u-1}+C=\frac{2}{1-\tan \frac{x}{2}}+C$$


\subsection{$\int \frac{2 x+4}{\left(x^{2}+4 x+5\right)^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{2 x+4}{\left(x^{2}+4 x+5\right)^{2}} \mathrm{~d} x=\int \frac{\mathrm{~d}\left(x^{2}+4 x+5\right)}{\left(x^{2}+4 x+5\right)^{2}}=-\frac{1}{\left(x^{2}+4 x+5\right)}+C$$


\subsection{$\int x \sin x^{2} \mathrm{~d} x$}
\textbf{解}\quad
$$\int x \sin x^{2} \mathrm{~d} x=\frac{1}{2} \int \sin x^{2} \mathrm{~d} x^{2}=\frac{-\cos x^{2}}{2}+C$$


\subsection{$\int \frac{1}{\sqrt[3]{5-3 x}} \mathrm{~d} x$;}
\textbf{解}\quad
$$\int(5-3 x)^{-\frac{1}{3}} \mathrm{~d} x=-\frac{1}{3}(5-3 x)^{-\frac{1}{3}} \mathrm{~d}(5-3 x)=-\frac{1}{2} (5-3 x)^{\frac{2}{3}}+C$$


\subsection{$\int \frac{1}{x \ln x \ln (\ln x)} \mathrm{~d} x$}
\textbf{解}\quad
$x>0$
$$\int \frac{1}{x \ln x \ln (\ln x)} \mathrm{~d} x=\int \frac{\mathrm{~d} \ln x}{\ln x \ln (\ln x)}=\int \frac{\mathrm{~d} \ln |(\ln x)|}{\ln (\ln x)}=\ln |(\ln  |(\ln x)|)|+C$$


\subsection{$\int \frac{1}{e^{x}+e^{-x}} \mathrm{~d} x$}
\textbf{解}\quad
令$u=e^x$

$$\int \frac{1}{\mathrm{e}^{x}+\mathrm{e}^{-x}} \mathrm{~d} x=\int \frac{e^{x}}{e^{2 x}+1} \mathrm{~d} x=\int \frac{\mathrm{~d} u}{u^{2}+1}=\arctan u+c=\arctan e^{x}+C$$


\subsection{$\int \sin 2 x \cos 3 x \mathrm{~d} x$}
\textbf{解}\quad
$$\int \sin 2 x \cos 3 x \mathrm{~d} x=\frac{1}{2} \int[\sin 5 x+\sin (-x)] \mathrm{~d} x=\frac{1}{2}\left[-\frac{1}{5} \cos 5x+\cos x\right]+C$$


\subsection{ $\int \frac{\arctan \sqrt{x}}{\sqrt{x}(1+x)} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{\arctan \sqrt{x}}{\sqrt{x}(1+x)} \mathrm{~d} x & =2 \int \frac{\arctan \sqrt{x}}{1+(\sqrt{x})^{2}} \mathrm{~d} \sqrt{x}             \\
		                                                          & =2 \int \operatorname{arctan} \sqrt{x} \mathrm{~d} \operatorname{\arctan} \sqrt{x} \\
		                                                          & =\arctan^2 \sqrt x +C
	\end{aligned}
$$

\subsection{ $\int \frac{1}{x \sqrt{1+x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
令$u=\sqrt{1+x^{2}}$,
$\mathrm{~d} x=\frac{\sqrt {x^2+1} }{x} \mathrm{~d} u$

$$
	\begin{aligned}
		\int \frac{1}{x \sqrt{1+x^{2}}} \mathrm{~d} x & =\int \frac{1}{x^{2}} \mathrm{~d} u                                                        \\
		                                              & =\int \frac{1}{u^{2}-1} \mathrm{~d} u                                                      \\
		                                              & = \frac{1}{2} \int \frac{1}{u-1} -\frac{1}{u+1} \mathrm{~d} u                              \\
		                                              & = \frac{1}{2}[\ln |u-1|-\ln |u+1|]                                                         \\
		                                              & = \frac{1}{2}\left[\ln \left|\sqrt{x^{2}+1}-1\right| -\left|\sqrt{x^{2}+1}+1\right|\right]
	\end{aligned}
$$


\subsection{ $\int \frac{1}{1+\sqrt{1-x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{1}{1+\sqrt{1-x^{2}}} \mathrm{~d} x & =\int \frac{1-\sqrt{1-x^{2}}}{x^{2}} \mathrm{~d} x                                                                                 \\
		                                              & =\int \frac{1}{x^{2}} \mathrm{~d} x-\int \frac{\sqrt{1-x^{2}}}{x^{2}} \mathrm{~d} x-\int \frac{\sqrt{1-x^{2}}}{x^{2}}\mathrm{~d} x \\
		                                              & =\frac{\sqrt{1-x^{2}}}{x}+\int \frac{1}{\sqrt{1-x^{2}}} \mathrm{~d} x                                                              \\
		                                              & =-\frac{1}{x} +\frac{\sqrt{1-x^{2}}}{x}+\arcsin x+C
	\end{aligned}
$$




\subsection{ $\int \frac{\sqrt{x^{2}-9}}{x} \mathrm{~d} x$}
\textbf{解}\quad
$u=\sqrt{x^{2}-9} \quad \mathrm{~d} u=\frac{x}{\sqrt{x^{2}-9}} \mathrm{~d} x \quad \mathrm{~d} x=\frac{\sqrt{x^{2}-9}}{x} \mathrm{~d} u$

$$\int \frac{\sqrt{x^{2}-9}}{x} \mathrm{~d} x=\int \frac{x^{2}-9}{x^{2}} \mathrm{~d} u=\int \frac{u^{2}}{u^{2}+9} \mathrm{~d} u=\int \mathrm{~d} u-\int \frac{9}{u^{2}+9}\mathrm{~d} u$$

$$\int \mathrm{~d} u = u +C$$

$$\int \frac{9}{u^{2}+9} \mathrm{~d} u=\int 3 \frac{1}{\left(\frac{u}{3}\right)^{2}+1} \mathrm{~d}\left(\frac{u}{3}\right)=3 \arctan \frac{u}{3}+C$$

$$\int \frac{\sqrt{x^{2}-9}}{x} \mathrm{~d} x=u-3 \arctan \frac{u}{3}+C=\sqrt{x^{2}-9}-3 \arctan \frac{\sqrt{x^{2}-9}}{3}+C$$


\subsection{ $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)^{3}}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)^{3}}} \mathrm{~d} x=\int \frac{a}{a^{3} \sqrt{\left(\frac{x^{2}}{a^{2}}+1\right)^{3}}} \mathrm{~d} \frac{x}{a}=\int \frac{1}{a^{2} \sqrt{\left(u^{2}+1\right)^{3}}} \mathrm{~d} u=\frac{1}{a^{2}} \int \frac{1}{\sqrt{\left(u^{2}+1\right)^{3}}} \mathrm{~d} u$$

令$u=\tan v$,则有
$$\frac{1}{a^{2}} \int \frac{1}{\sqrt{\left(u^{2}+1\right)^{3}}} \mathrm{~d} u=\frac{1}{a^{2}} \int \cos v \mathrm{~d} u=\frac{\sin v}{a^{2}}+C=\frac{1}{a^{2}} \sin \arctan \frac{x}{a}+C=\frac{x}{a^{2} \sqrt{a^{2}+x^{2}}}+C$$

最后一步可由$\mathrm{Rt}$三角型的性质得出.


\subsection{ $\int \sqrt{5-4 x-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \sqrt{5-4 x-x^{2}} \mathrm{~d} x=\int \sqrt{9-(x+2)^{2}} \mathrm{~d}(x+2)=\int \sqrt{9-t^{2}} \mathrm{~d} t$$

令$t=3\sin v$

$$
	\begin{aligned}
		9\int\cos ^{2} v \mathrm{~d} v & =9\int \frac{1+ \cos 2v}{2} \mathrm{~d} v \\&=\frac{9v}{2}+\frac{9}{4}\sin 2v+C\\&=\frac{9}{2}\arcsin \left(\frac{x+2}{3}\right)+\frac{9}{4} \sin \left(2 \arcsin \left(\frac{x+2}{3}\right)\right)+C
	\end{aligned}
$$

\section{2.应用分部积分法求下列不定积分 :}
\subsection{$\int \arcsin x \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \arcsin x \mathrm{~d} x & =x \arcsin x-\int x \frac{1}{\sqrt{1-x^{2}}} \mathrm{~d} x \\&=x \arcsin x-\frac{1}{2} \int \frac{\mathrm{~d} x^{2}}{\sqrt{1-x^{2}}} \\&=x \arcsin x+\sqrt{1-x^{2}}+C
	\end{aligned}
$$

\subsection{$\int \ln (1+x) \mathrm{~d} x$}
\textbf{解}\quad
$x+1>0$

$$
	\begin{aligned}
		\int \ln (1+x) \mathrm{~d} x & =x \ln (1+x)-\int \frac{x}{1+x} \mathrm{~d} x \\&=x \ln (1+x)-\int \mathrm{~d} x+\int \frac{\mathrm{~d} x}{1+x}\\&=x \ln (1+x)-x+\ln (x+1)+C
	\end{aligned}
$$


\subsection{$\int \frac{x^{2}}{1+x^{2}} \arctan x \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{x^{2}}{1+x^{2}} \arctan x \mathrm{~d} x & =\int\arctan x \mathrm{~d} x-\int \frac{1}{1+x^{2}}  \arctan x\mathrm{~d} x \\&=\int \arctan x \mathrm{~d} x-\int \arctan x \mathrm{~d} \arctan x\\&=x \arctan x-\int \frac{x}{1+x^{2}} \mathrm{~d} x-\frac{1}{2} \arctan ^{2} x\\&=x \arctan x-\frac{1}{2} \arctan ^{2} x-\frac{1}{2} \ln \left(1+x^{2}\right)+C
	\end{aligned}
$$


\subsection{$\int x^{2} \sin x \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int x^2 \sin x \mathrm{~d} x & = -x^2 \cos x + \int  2x \cos x \mathrm{~d} x \\&= -x^2 \cos x + 2 \int x \cos x \mathrm{~d} x \\&= -x^2 +2\left( x \sin x - \int \ sin x \mathrm{~d} x  \right)\\&=-x^2 \cos x + 2 x \sin x +2 \cos x +C
	\end{aligned}
$$

\subsection{$\int x^{2} \mathrm{e}^{-x} \mathrm{~d} x$}
\textbf{解}\quad
$\int x^{2} e^{-x} \mathrm{~d} x=-e^{-x} x^{2}+2 \int x e^{-x} \mathrm{~d} x=-x^{2} e^{-x}+2\left(x e^{-x}+\int e^{-x} \mathrm{~d} x\right)=-x^{2} e^{-x}-2 x e^{-x}-2 e^{-x}+C$


\subsection{$\int e^{\sqrt[3]{x}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int e^{\sqrt[3]{x}} \mathrm{~d} x & =\int e^{t} \mathrm{~d} t^{3} \\&=3 \int t^{2} e^{t} \mathrm{~d} t\\&=3\left(t^{2} e^{t}-2 \int t e^{t}\right)\\&=3 t^{2} e^{t}-6 \int t e^{t}\\&=3 t^{2} e^{t}-b\left(t e^{t}-\int e^{t} \mathrm{~d} t\right)\\&=3 t^{2} t^{t}-6 t e^{t}+t e^{t}\\&=3 \sqrt[3]{x^2} e^{\sqrt[3]{x}}-6 \sqrt[3]{x} e^{3 \sqrt{x}}+6 e^{\sqrt[3]{x}}+C
	\end{aligned}
$$


\subsection{$\int \frac{1}{1+\sqrt[3]{x+1}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{1}{1+\sqrt[3]{x+1}} \mathrm{~d} x & =\int \frac{1}{1+\sqrt[3]{t}} \mathrm{~d} t \\&=\int \frac{1}{1+u} \mathrm{~d} u^{3}\\&=\int \frac{3 u^{2}}{1+u} \mathrm{~d} u\\&=\int(3 u-3) \mathrm{~d} u+\int \frac{3}{1+u} \mathrm{~d} u\\&=\frac{3}{2} u^{2}-3 u+3 \ln |1+u|+C\\&=\frac{3}{2} \sqrt[3]{(x+1)^{2}}-3 \sqrt[3]{x+1}+3 \ln (1+\sqrt[3]{x+1})+C
	\end{aligned}
$$


\subsection{$\int \frac{1}{x\left(2+x^{10}\right)} \mathrm{~d} x$}
\textbf{解}\quad
令$u=x^{10}$,则$\mathrm{~d} u = 10x^9\mathrm{~d} x,\mathrm{~d} x=\frac{\mathrm{~d} u}{10x^{9}}$

$$
	\begin{aligned}
		\int \frac{1}{x\left(2+x^{10}\right)} \mathrm{~d} x & =\int \frac{1}{10 u(2+u)} \mathrm{~d} u \\&=\int \frac{1}{10}\left(\frac{1}{2} \int\frac{1}{u} \mathrm{~d} u-\frac{1}{2} \int \frac{1}{u+2} \mathrm{~d} u\right)\\&=\frac{1}{20} \ln \frac{x^{10}}{x^{10}+2}+C
	\end{aligned}
$$


\subsection{$\int \frac{1}{x \sqrt{1+x^{4}}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{1}{x \sqrt{1+x^{4}}} \mathrm{~d} x & =\frac{1}{2} \int \frac{1}{x^{2} \sqrt{1+x^{4}}} \mathrm{~d} x^{2} \\&=\frac{1}{2} \int \frac{\mathrm{~d} t}{t \sqrt{1+t^{2}}}\\&=\frac{1}{2}\int \frac{\mathrm{~d} \sqrt{t^{2}+1}}{\left(\sqrt{t^{2}+1}\right)^{2}-1}  \\&=\frac{1}{4} \ln \frac{\sqrt{x^{4}+1}-1}{\sqrt{x^{4}+1}+1}+C
	\end{aligned}
$$

\section{已知 $\frac{\sin x}{x}$ 是 $f(x)$ 的原函数,求 $\int x f^{\prime}(x) \mathrm{~d} x$.}
\textbf{解}\quad
$$\int x f^{\prime}(x) \mathrm{~d} x=x f(x)-\int f(x) \mathrm{~d} x$$

$$\because \int f(x) \mathrm{~d} x=\frac{\sin x}{x} \therefore f(x)=\left(\frac{\sin x}{x}\right)^{\prime}=\frac{x \cos x-\sin x}{x^{2}}$$

$$\therefore \int x f^{\prime}(x) \mathrm{~d} x=\frac{x \cos x-\sin x}{x}-\frac{\sin x}{x}+C=\cos x-\frac{2 \sin x}{x}+C$$

\section{计算下列不定积分 ($a, b$ 均为非零常数)}
\subsection{$\int \frac{x}{a x+b} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{x}{a x+b} \mathrm {\mathrm{~d}} x=\int \frac{1}{a} \mathrm{~d} x-\int \frac{\frac{b}{a}}{a x+b} \mathrm{~d} x=\int \frac{1}{a} \mathrm{~d} x-\frac{b}{a^{2}} \int \frac{1}{a x+b}\mathrm{~d} (a x+b)=\frac{x}{a}-\frac{b}{a^{2}} \ln |a x+b|+C$$

注：积分出来的常数项如果真的是个常数，可以把它和$\mathrm{C}$合并同类项，而在重积分中，用主元法的时候认为其它变量都是常数，但如果将其带其它变量的常数项归为$\mathrm{C}$，则一定是错的。


\subsection{$\int \frac{x^{2}}{a x+b} \mathrm{~d} x$}
\textbf{解}\quad
令$u=ax+b$,则$\mathrm{~d}x=\frac{\mathrm{~d} u}{a}$

$$
	\begin{aligned}
		\int \frac{x^{2}}{a x+b} \mathrm{~d} x & =\int \frac{\left(\frac{u-b}{a}\right)^{2}}{a u} \mathrm{~d} u \\&=\frac{1}{a^{3}} \int \frac{(n-b)^{2}}{u}\mathrm{~d}x\\&=\frac{1}{a^{2}}\left(\int u \mathrm{~d} u-2 b \int \mathrm{~d} u+\frac{b^{2}}{u} \mathrm{~d} u\right)\\&=\frac{1}{a^{3}}\left(\frac{u^{2}}{2}-2 b u+b^{2} \ln |u|\right)+C\\&=\frac{1}{2 a^{3}}\left(a^{2} x^{2}-2 a b x+2 b^{2} \ln |a x+b|-3 b^{2}\right)+C\\&=\frac{x^{2}}{2 a}-\frac{b}{a^{2}} x+\frac{b^{2}}{a^{3}} \ln |a x+b|+C
	\end{aligned}
$$


\subsection{$\int \frac{1}{x(a x+b)} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{1}{x(a x+b)} \mathrm{~d} x & =\frac{1}{a} \int \frac{1}{x\left(x+\frac{b}{a}\right)} \mathrm{~d} x \\&=\frac{1}{b}\int\left(\frac{1}{x}-\frac{1}{x+\frac{b}{a}}\right) \mathrm{~d} x\\&=\frac{1}{b} \ln \left|\frac{x}{x+\frac{b}{a}}\right|+C\\&=\frac{1}{b} \ln \left|\frac{a x}{a x+b}\right|+C
	\end{aligned}
$$


\subsection{$\int \frac{x}{(a x+b)^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \frac{x}{(a x+b)^{2}} \mathrm{~d} x & =\frac{1}{a} \int \frac{a x}{(a x+b)^{2}} \mathrm{~d} x \\&=\frac{1}{a^{2}} \int \frac{a x+b-b}{(a x+b)^{2}} \mathrm{~d}(a x+b)\\&=\frac{1}{a^{2}} \int \frac{u-b}{u^{2}} \mathrm{~d} u\\&=\frac{1}{a^{2}}\left(\int \frac{\mathrm{~d} u}{u}-\int \frac{b}{u^{2}} \mathrm{~d} u\right)\\&=\frac{1}{a^{2}}\left((\ln|a x+b|+\frac{b}{a x+b}\right)+C
	\end{aligned}
$$



\subsection{$\int \sqrt{a x+b} \mathrm{~d} x$}
\textbf{解}\quad
$$\int(a x+b)^{\frac{1}{2}} \mathrm{~d} x=\frac{1}{a} \int(a x+b)^{\frac{1}{2}} \mathrm{~d}(a x+b)=\frac{1}{a} \int u^{\frac{1}{2}} \mathrm{~d} u=\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+C$$


\subsection{$\int x \sqrt{a x+b} \mathrm{~d} x$}
\textbf{解}\quad
令$u=\sqrt{ax+b} $，则$\mathrm{~d} x=\frac{\sqrt{a x+b}}{a} \mathrm{~d} u$

$$
	\begin{aligned}
		\int x \sqrt{a x+b} \mathrm{~d} x & =\int \frac{u^{2}-b}{a} \cdot u \cdot \frac{2 u}{a} \mathrm{~d} u \\&=\frac{2}{a^{2}} \int u^{2}\left(h^{2}-b\right) \mathrm{~d} u\\&=\frac{2}{a^{2}}\left(\frac{1}{5} u^{5}-\frac{b}{3} u^{3}\right)+C\\&=\frac{2}{a^{2}}\left(\frac{1}{5}(a x+b)^{\frac{5}{2}}-\frac{b}{3}(a x+b)^{\frac{3}{2}}\right)+C
	\end{aligned}
$$


\subsection{$\int \frac{x^{2}}{\sqrt{x^{2}+a^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
令$x=a \tan u$，则$\mathrm{~d} x = a \sec^2 u \mathrm{~d} u$

$$
	\begin{aligned}
		\int \frac{x^{2}}{\sqrt{x^{2}+a^{2}}} \mathrm{~d} x & =\int \frac{(a \tan u)^{2}}{\operatorname{a sec} u}\operatorname{asec}^{2} u \mathrm{~d} u \\&=\int a^{2} \tan ^{2} u \sec u \mathrm{~d} u\\&=a^{2} \int\left(\sec ^{2} u-1\right) \sec u \mathrm{~d} u\\&=a^{2}\left(\int \sec ^{3} u \mathrm{~d} u-\int \sec u \mathrm{~d} u\right)
	\end{aligned}
$$

$$
	\begin{aligned}
		\int \sec u \mathrm{~d} u & =\int \frac{\cos u}{\cos ^{2} u} \mathrm{~d} u \\&=\int \frac{\mathrm{~d} \sin u}{1-\sin ^{2} u}\\&=\frac{1}{2} \int \frac{1}{\sin u+1} \mathrm{~d} u-\frac{1}{2} \int \frac{1}{\sin u-1} \mathrm{~d} u\\&=\frac{1}{2} \ln \left|\frac{1+\sin u}{1-\sin u}\right|+C\\&=\ln \sqrt{\frac{(\sin u+1)^{2}}{\cos ^{2} u}}\\&=\ln |\tan x+\sec x|+C
	\end{aligned}
$$

$$
	\begin{aligned}
		\int \sec ^{3} u \mathrm{~d} u & =\int \frac{\mathrm{~d} u}{\cos ^{3} u}                           \\&=\sec u \tan u-\int \frac{\sin ^{2} u}{\cos ^{3} u} \mathrm{~d} u\\&=\sec u \tan u-\int \frac{1}{\cos ^{3} u} \mathrm{~d} u+\int \frac{\mathrm{~d} u}{\cos u}\\&=\sec u \tan u-\int \sec ^{3} u \mathrm{~d} u+\sec u \mathrm{~d} u \\
		                               & =\sec u \tan u+\ln |\tan u+\sec u|-\int \sec ^{3} u \mathrm{~d} u
	\end{aligned}
$$

$$\therefore \int \sec ^{3} u \mathrm{~d} u=\frac{1}{2}(\sec u \tan u+\ln |\mathrm{~d} \tan u+\sec u |)+C$$

$$\begin{aligned}
		\int \frac{x^{2}}{\sqrt{x^{2}+a^{2}}} \mathrm{~d} x & =a^{2}\left(\int \sec ^{3} u \mathrm{~d} u-\int \sec u \mathrm{~d} u\right)                    \\
		                                                    & =a^{2}\left(\frac{\sec u \tan n}{2}+\frac{\ln |\tan u+\sec u|}{2}-\ln |\tan u+\sec u|\right)+C \\ & =a^{2}\left(\frac{\sec u \tan u}{2}-\frac{\ln |\tan u+\sec u|}{2}\right)+C \\& =\frac{a^{2}}{2}\left(\frac{\sqrt{a^{2}+x^{2}}}{a} \cdot \frac{x}{a}-\ln \left|\frac{x}{a}+\frac{\sqrt{x^{2}+a^{2}}}{a}\right|\right)+C\\&=\frac{a^{2}}{2}\left(\frac{x \sqrt{a^{2}+x^{2}}}{a^{2}}-\ln \left|x+\sqrt{x^{2}+a^{2}}\right|\right)+C \\& =\frac{x \sqrt{a^{2}+x^{2}}}{2}-\frac{a^{2}}{2} \ln \left|x+\sqrt{x^{2}+a^{2}}\right|+C\end{aligned}$$


\subsection{$\int \sqrt{\left(x^{2}+a^{2}\right)^{3}} \mathrm{~d} x$}
\textbf{解}\quad
令$\tan u = \frac{x}{a}$，有

$$
	\begin{aligned}
		\int \sqrt{\left(x^{2}+a^{2}\right)^{3}} \mathrm{~d} x & =a^4 \int\sec ^{5} u \mathrm{~d} u \\&=a^4 \sec ^{3} u \tan u-3 a^4 \int \tan^2 u \sec ^{3} u \mathrm{~d} u\\&=a ^4\sec ^{3} u \tan u-3 a^4\left(\int \sec ^{5} u \mathrm{~d} u-\int \sec ^{3} u \mathrm{~d} u\right)
	\end{aligned}
$$

$$\therefore \int \sec ^{5} u \mathrm{~d} u=\frac{1}{4} \sec ^{3} u \cdot \tan u+\frac{3}{4} \int \sec ^{3} u \mathrm{~d} u$$

根据上题结论，

$$\int \sec ^{3} u \mathrm{~d} u=\frac{1}{2}(\sec u \tan u+\ln |\tan u+\sec u|)$$

$$
	\begin{aligned}
		\therefore \int\sec ^{5} u \mathrm{~d} u & =\frac{\sec^3u\tan u}{4}+\frac{3}{8}\sec u \tan u + \frac{3}{8} \ln |\tan u +\sec u| \\&=\frac{x \sqrt{\left(x^{2}+a^{2}\right)^{3}}}{4 a^{4}}+\frac{3 x \sqrt{x^{2}+a^{2}}}{8 a^{2}}+\frac{3}{8} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)+C
	\end{aligned}
$$

$$\therefore  \int \sqrt{\left(x^{2}+a^{2}\right)^{3}} \mathrm{~d} x=\frac{x \sqrt{\left(x^{2}+a^{2}\right)^{3}}}{4}+\frac{3a^{2} x \sqrt{x^{2}+a^{2}}}{8 }+\frac{a^{4}3}{8} \ln \left(x+\sqrt{x^{2}+a^{2}}\right)+C$$


\subsection{$\int x \arcsin \frac{x}{a} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int x \arcsin \frac{x}{a} \mathrm{~d} x & =a^{2} \int \frac{x}{a} \operatorname{ancsin} \frac{x}{a} \mathrm{~d} \frac{x}{a} \\&=a^{2} \int u \arcsin u \mathrm{~d} u\\&=a^{2}\left(\frac{1}{2} u^{2} \arcsin u-\frac{1}{2} \int \frac{u^{2}}{\sqrt{1-u^{2}}} \mathrm{~d} u\right)\\&=\frac{a^{2}}{2} u^{2} \arcsin u-\frac{a^{2}}{4}\left(\arcsin u-\frac{1}{2} \sin \left(2 \arcsin u\right)\right)+C\\&=\frac{1}{2} x^{2} \arcsin \frac{x}{a}-\frac{a^{2}}{4} \arcsin \frac{x}{a}+\frac{a^{2}}{8} \sin \left(2 \arcsin \frac{x}{a}\right)+C
	\end{aligned}
$$


\subsection{$\int e^{a x} \sin b x \mathrm{~d} x$}
\textbf{解}\quad
若 $a=b=0,$ 则积分为 $x+C$; 若 $a=0, b \neq 0,$ 则积分为 $-\frac{1}{b} \cos b x+C$; 以下设 $a \neq 0$.
$$
	\begin{aligned}{l}
		\int \mathrm{e}^{a x} \sin b x \mathrm{~d} x & =\frac{1}{a} \int \sin b x \mathrm{~d}\left(\mathrm{e}^{a x}\right) \\&=\frac{1}{a} \mathrm{e}^{a x} \sin b x-\frac{b}{a} \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x\\&=\frac{1}{a} \mathrm{e}^{a x} \sin b x-\frac{b}{a^{2}} \int \cos b x \mathrm{~d}\left(\mathrm{e}^{a x}\right) \\&=\frac{1}{a} \mathrm{e}^{a x} \sin b x-\frac{b}{a^{2}} \mathrm{e}^{a x} \cos b x-\frac{b^{2}}{a^{2}} \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x
	\end{aligned}
$$

于是,得 $$\int \mathrm{e}^{a x} \sin b x \mathrm{~d} x=\frac{\mathrm{e}^{a x}(a \sin b x-b \cos b x)}{a^{2}+b^{2}}+C$$


\subsection{$\int(\arcsin x)^{2} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int \operatorname{arcsin}^{2} x \mathrm{~d} x & =x \arcsin ^{2} x-\int \frac{2 x \operatorname{ancsin} x}{\sqrt{1-x^{2}}} \mathrm{~d} x \\\int \frac{2 x \arcsin x}{\sqrt{1-x^{2}}} \mathrm{~d} x&=2 \int x \arcsin x \frac{1}{\sqrt{1-x^{2}}} \mathrm{~d} x\\&=2\left(-\arcsin x \sqrt{1-x^{2}}-(-x)\right)\\&=2\left(-\arcsin x \sqrt{1-x^{2}}+x\right)
	\end{aligned}
$$

$$\int \arcsin ^{2} x \mathrm{~d} x=x \arcsin ^{2} x+2 \arcsin x \sqrt{1-x^{2}}-2 x+C$$

令 $\arcsin x=t,$ 则

$$\begin{aligned} \int(\arcsin x)^{2} \mathrm{~d} x &=\int t^{2} \cos t \mathrm{~d} t\\&=t^{2} \sin t-2 \int t \sin t \mathrm{~d} t\\&=t^{2} \sin t+2 t \cos t-2\sin t \\ &=(\arcsin x)^{2} x+2 \arcsin x \sqrt{1-x^{2}}-2x+C \end{aligned}$$


\subsection{$\int \mathrm{e}^{\sqrt{a x+b}} \mathrm{~d} x$}
\textbf{解}\quad
令$u=ax+b$

$$
	\begin{aligned}
		\int e^{\sqrt{a x+b}} \mathrm{~d} x & =\int \frac{e^{\sqrt{u}}}{a} \mathrm{~d} u=\frac{1}{a} \int e^{\sqrt{u}} \mathrm{~d} u=\frac{1}{a} \int e^{v} 2 v \mathrm{~d} v \\&=\frac{2}{a}\left(e^{v} v-\int e^{v} \mathrm{~d} v\right)=\frac{2}{a}\left(e^{v} v-e^{v}\right)\\&=\frac{2}{a}(\sqrt{a x+b}-1) e^{\sqrt{a x+b}}
	\end{aligned}
$$



\subsection{$\int \frac{1}{\sqrt{x-x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
$$\int \frac{1}{\sqrt{x-x^{2}}} \mathrm{~d} x=\int \frac{1}{\sqrt{\frac{1}{4}-\left(x+\frac{1}{2}\right)^{2}}} \mathrm{~d} x= \int \frac{1}{\sqrt{1-(2 x+1)^{2}}} \mathrm{~d}(2 x+1)=\arcsin (2x+1)+C$$


\subsection{$\int \sqrt{\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{1+x^{2}}} \mathrm{~d} x$}
\textbf{解}\quad
令$u=x+\sqrt {1+x^2}$,$\mathrm{~d} x=\frac{\sqrt{1+x^{2}}}{x+\sqrt{1+x^{2}}} \mathrm{~d} u$
$$
	\begin{aligned}
		\int \sqrt{\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{1+x^{2}}} \mathrm{~d} x & =\int \sqrt{\frac{\ln u}{1+x^{2}}} \cdot \sqrt{\frac{1+x^{2}}{u^{2}}} \mathrm{~d} u \\&=\int \frac{\sqrt{\ln u}}{u} \mathrm{~d} u=\int \sqrt{\ln u} \mathrm{~d} \ln u=\int \sqrt{v} \mathrm{~d} v\\&=\frac{2}{3} v^{\frac{3}{2}}
	\end{aligned}
$$

逐步带回，得
$$\int \sqrt{\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{1+x^{2}}} \mathrm{~d} x =\frac{2}{3}\left[\ln \left(x+\sqrt{1+x^{2}}\right)\right]^{\frac{3}{2}}+C$$

\subsection{$\int \frac{1}{\sin x \cos ^{3} x} \mathrm{~d} x$}
\textbf{解}\quad
令$x=\arctan u$,$\mathrm{~d} x=\frac{1}{1+u^{2}} \mathrm{~d} u$

$$
	\begin{aligned}
		\int \frac{1}{\sin x \cos ^{3} x} \mathrm{~d} x & =\int \frac{1}{\frac{u}{\sqrt{u^{2}+1}} \frac{1}{\left(\sqrt{u^{2}+1}\right)^{3}} } \mathrm{~d} x \\&=\int \frac{u^{2}+1}{u} \mathrm{~d} u\\&=\int u \mathrm{~d} u+\int \frac{\mathrm{~d} u}{u}\\&=\frac{u^{2}}{2}+\ln |u|+C\\&=\frac{\tan ^{2} x}{2}+\ln |\tan x|+C
	\end{aligned}
$$

\subsection{$\int \frac{1}{\sqrt{1+\mathrm{e}^{x}}} \mathrm{~d} x$}
\textbf{解}\quad
$\int \frac{1}{\sqrt{1+e^{x}}} \mathrm{~d} x=\int \frac{1}{\sqrt{1+e^{x}}} \cdot \frac{1}{e^{x}} \cdot \mathrm{~d} e^{x}=\int \frac{1}{\sqrt{1+u}} \cdot \frac{1}{u} \mathrm{~d} u$

$$
	\begin{aligned}
		\int \frac{1}{u \sqrt{1+u}} \mathrm{~d} u & =\int \frac{2}{v^{2}-1} \mathrm{~d} v \\&=\int\left(\frac{1}{v-1}-\frac{1}{v+1}\right) \mathrm{~d} v\\&=\ln \left|\frac{v-1}{v+1}\right|+C\\&=\ln \left|\frac{\sqrt{1+u}-1}{\sqrt{1+u}+1}\right|+C\\&=\ln \left|\frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}\right|+C
	\end{aligned}
$$


\subsection{$\int x \sqrt[3]{1-x} \mathrm{~d} x$}
\textbf{解}\quad
令$\sqrt[3]{1-x}=u$，

$$\int x \sqrt[3]{1-x} \mathrm{~d} x=\int -3 u^{3}\left(1-u^{3}\right) \mathrm{~d} u=3 \int u^{6}-u^{3} \mathrm{~d} u=\frac{3}{7} u^{7}-\frac{3}{4} u^{4}+C=\frac{3}{7}(1-x)^{\frac{7}{3}}-\frac{3}{4}(1-x)^{\frac{4}{3}}+C$$


\subsection{$\int \sqrt{x} \ln ^{2} x \mathrm{~d} x$}
\textbf{解}\quad
$$\int \sqrt{x} \ln ^{2} x \mathrm{~d} x=\frac{2}{3} x^{\frac{3}{2}} \ln ^{2} x-\frac{4}{3} \int \sqrt{x} \ln x \mathrm{~d} x$$

$$\int \sqrt{x} \ln x \mathrm{~d} x=\frac{2}{3} x^{\frac{3}{2}} \ln x-\int \frac{2}{3} \sqrt{x} \mathrm{~d} x=\frac{2}{3} x^{\frac{3}{2}} \ln x-\frac{4}{9} x^{\frac{3}{2}}$$

$$\int \sqrt{x} \ln ^{2} x \mathrm{~d} x=\frac{2}{3} x^{\frac{3}{2}} \ln ^{2} x-\frac{8}{9} x^{\frac{3}{2}} \ln x+\frac{16}{27} x^{\frac{3}{2}}+C$$

\subsection{$\int x^{2} \arccos x \mathrm{~d} x$}
\textbf{解}\quad
$$\int x^{2} \arccos x \mathrm{~d} x=\frac{x^{3}}{3} \arccos x+\int \frac{x^{3}}{3 \sqrt{1-x^{2}}} \mathrm{~d} x$$

$$\int \frac{x^{3}}{3 \sqrt{1-x^{2}}} \mathrm{~d} x=\frac{1}{3} \int \frac{x^{3}}{\sqrt{1-x^{2}}} \mathrm{~d} x$$

令$u=\sqrt {1-x^2}$，得

$$
	\mathrm{LHS}=\frac{1}{3} \int u^{2} \mathrm{~d} u-\frac{1}{3} \int \mathrm{~d} u= \frac{u^{3}}{9}-\frac{u}{3}+C
$$

$$
	\therefore \int x^{2} \arccos x \mathrm{~d} x=\frac{x^{3}}{3} \arccos x+\frac{\left(\sqrt{1-x^{2}}\right)^{3}}{9}-\frac{\sqrt{1-x^{2}}}{3}+C
$$


\subsection{$\int x \cos \sqrt{x} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\int x \cos \sqrt{x} \mathrm{~d} x & =\int 2 x \sqrt{x} \cos \sqrt{x} \mathrm{~d} \sqrt{x} \\&=2 \int u^{3} \cos u \mathrm{~d} u\\&=2 \int u^{3} \mathrm{~d} \sin u=2 u^{3} \sin u+6\left(\int u^{2}  \mathrm{~d} \cos u\right)
	\end{aligned}
$$

$$\int u^{2} \mathrm{~d} \cos u=u^{2} \cos u-2 \int u \cos u \mathrm{~d} u=u^{2} \cos u-2 \int u \mathrm{~d} \sin u$$

$$\int u \mathrm{~d} \sin n=u \sin u-\int \sin u \mathrm{~d} u=u \sin u +\cos u+C$$

$$\therefore \int u^{2} \mathrm{~d} \cos u=u^{2} \cos u-2 u \sin u-2 \cos u+C$$

$$
	\begin{aligned}
		\therefore \int x \cos \sqrt{x} & =2 \int u^{3} \mathrm{~d} \sin u \\&=2 u^{3} \sin u+6 u^{2} \cos u-12 u \sin u-12 \cos u+C\\&=2 x^{\frac{3}{2}} \sin \sqrt{x}+6 x \cos \sqrt{x}-12 \sqrt{x} \sin \sqrt{x}-12 \cos \sqrt{x}+C
	\end{aligned}
$$

\section{证明下列命题:}
\subsection{设 $I_{n}=\int \tan ^{n} x \mathrm{~d} x,(n>1),$ 求证 $: I_{n}=\frac{1}{n-1} \tan ^{n-1} x-I_{n-2},$ 并求 $\int \tan ^{5} x \mathrm{~d} x$}
\textbf{证}\quad
利用 $\tan ^{2} x=\sec ^{2} x-1,$ 有
$$
	\begin{aligned}
		I_{n} & =\int \tan ^{n-2} x \frac{1-\cos ^{2} x}{\cos ^{2} x} \mathrm{~d} x     \\
		      & =\int \tan ^{n-2} x \mathrm{~d} \tan x-\int \tan ^{n-2} x \mathrm{~d} x \\
		      & =\frac{1}{n-1} \tan ^{n-1} x-I_{n-2}
	\end{aligned}
$$

\subsection{设 $I_{n}=\int \frac{1}{\left(x^{2}+a^{2}\right)^{n}} \mathrm{~d} x,(n>1),$ 求证: $I_{n}=\frac{2 n-3}{2 a^{2}(n-1)} I_{n-1}+\frac{1}{2 a^{2}(n-1)}\cdot \frac{x}{\left(x^{2}+a^{2}\right)^{n-1}}$}
\textbf{证}\quad
$$I_{n}=\int \frac{1}{\left(x^{2}+a^{2}\right)^{n}} \mathrm{~d} x=\int \frac{\mathrm{~d} x}{\left(\frac{x^{2}}{a^{2}}+1\right)^{n} \cdot a^{2 n}}=\frac{1}{a^{2 n}} \int \frac{\mathrm{~d} x}{\left(\frac{x^{2}}{a^{2}}+1\right)^{n}}$$

 令$\tan t=\frac{x}{a}, x=a \tan t$
$$
\begin{aligned} I_{n} &=\frac{1}{a^{2 n}} \int \frac{a \sec ^{2} t}{\left(\sec ^{2} t\right)^{n}} \mathrm{~d} t=\frac{1}{a^{2 n-1}} \int \frac{1}{\left(\sec ^{2} t\right)^{n-1}} \mathrm{~d} t \\ &=\frac{1}{a^{2 n-1}} \int\left(\cos ^{2} t\right)^{n-1} \mathrm{~d} t=\frac{1}{a^{2 n-1}} \int \cos ^{2 n-2} t \mathrm{~d} t \end{aligned}
$$
$$
\begin{aligned}
\int \cos ^{n} t d t&=\int \cos ^{n-1} t d \sin t=\cos ^{n-1} t \sin t+(n-1) \int \cos ^{n-2} t \sin ^{2} t d t\\&=\cos ^{n-1} t \sin t+(n-1) \int \cos ^{n-2} t (1-\cos^2t)  d t
\\&
=\cos ^{n-1} t \sin t+(n-1) \int \cos ^{n-2} t d t-(n-1) \int \cos ^{n} t d t
\end{aligned}
$$

$$\therefore n \int \cos ^{n} t \mathrm{~d} t=\cos ^{n-1} t \sin t+(n-1) \int \cos ^{n-2} t \mathrm{~d} t$$

$$\therefore \int \cos ^{n} t \mathrm{~d} t=\frac{\cos ^{n-1} t \sin t}{n}+\frac{n-1}{n} \int \cos ^{n-2} t \mathrm{~d} t$$

$$
\begin{aligned}
I_{n}&=\frac{1}{a^{2 n-1}} \int \cos ^{2 n-2} t d t\\&=\frac{1}{a^{2 n-1}(2n-2)} \cos ^{2 n-3} x \sin x+\frac{1}{a^{2 n-1}} \cdot \frac{2 n-3}{2 n-2} \int \cos ^{2 n-4} t d t\\&=\frac{1}{a^{2 n-1}(2n-2)} \cos ^{2 n-3} t \sin t+\frac{1}{a^{2 n-1}} \cdot \frac{2 n-3}{2 n-2} a^{2n-3}I_n\\&
=\frac{1}{a^{2 n-1}(2n-2)}\left(\frac{a}{\sqrt{x^{2}+a^{2}}}\right)^{2 n-3}\left(\frac{x}{\sqrt{x^{2}+a^{2}}}\right)+\frac{2 n-3}{2 a^{2}(n-1)} I_{n-1}
\\&=\frac{1}{2 a^{2}(n-1)} \cdot \frac{x}{\left(x^{2}+a^{2}\right)^{n-1}}+\frac{2 n-3}{2 a^{2}(n-1)}I_{n-1}
\end{aligned}
$$

证毕。








\end{document}
\subsection{}
\textbf{解}\quad

\subsection{}
\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad